Note from (REF ) and (REF ) that
\(\mathcal {K}(x,y)=(4\pi )^{-d/2}\,\int _0^\infty e^{-tR}e^{-\frac{|x-y|^2}{4t}}\,\frac{dt}{t^{1+\sigma +\frac{d}{2}}}.\) 
For \(\nu >0\) , consider the special function
\(F_\nu (z):=\int _0^\infty e^{-s}e^{-\frac{z^2}{4s}}\,s^{\nu -1}\,ds\,\lesssim (1+z)^{\nu -\frac{1}{2}}\,e^{-z},\quad z>0,\) 
where the inequality follows from the asymptotics of the integral; see e.g. [1]}.
If we change variables \(s=|x-y|^2/(4t)\)  in (REF ) we obtain that
\(\mathcal {K}(x,y) = \,c \,\frac{F_{\sigma + d/2}\big (\sqrt{R}\,|x-y|\big )}{|x-y|^{2\sigma +d}}\,\lesssim \, \frac{(1+\sqrt{R}\,|x-y|)^{\sigma +\frac{d-1}{2}}}{|x-y|^{2\sigma +d}}\,e^{-\sqrt{R}\,|x-y|}.\) 
If we now assume that \(|y|\ge 2(|x|\vee 1)\) , the right side is easily seen to be controlled by
\(c(x)\,\Phi _\sigma (y)\) ; see e.g. [2]} and subsequent lines for a detailed argument.
